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**cakeboby** Question: Let a and b be the roots of the quadratic equation $\displaystyle x^2- (m +\frac 3 m )x +2 = 0$ where m is a non-zero constant.

Using the fact that $\displaystyle (a-b)^2 \geq 0$ for all real numbers a and b, show that $\displaystyle m^2 +\frac 9 {m}^2 \geq 6$.

Hence show tha$\displaystyle t (m + \frac 3 m }^2 \geq 12$

MY WORK:

$\displaystyle (a-b)^2 = (a+b)^2 - 4ab = (a+b)^2 - 8 \geq 0$

$\displaystyle (m+ \frac 3 m ) -8 \geq 0$

$\displaystyle m^2+6m+\frac 9 {m}^2 \geq 8 $

Thanks in advance.